Enter steady flow inputs
Use the common sign convention: heat into the control volume is positive, and shaft work delivered by the control volume is positive.
Example data table
This sample demonstrates a turbine-style control volume with measurable enthalpy drop and modest kinetic and potential energy changes.
| Case | ṁ (kg/s) | Q̇ (kW) | Ẇ (kW) | h₁ (kJ/kg) | h₂ (kJ/kg) | V₁ (m/s) | V₂ (m/s) | z₁ (m) | z₂ (m) |
|---|---|---|---|---|---|---|---|---|---|
| Steam Turbine A | 2.50 | 150.00 | 1,509.68 | 3,250.00 | 2,700.00 | 45.00 | 120.00 | 8.00 | 2.00 |
| Nozzle B | 1.80 | 0.01 | 0.00 | 3,050.00 | 2,940.00 | 30.00 | 470.00 | 1.00 | 1.50 |
| Compressor C | 3.10 | -18.00 | -329.82 | 280.00 | 380.00 | 22.00 | 40.00 | 0.00 | 3.00 |
Formula used
For a one-inlet, one-outlet steady flow device, the steady flow energy equation balances heat transfer, shaft work, and changes in specific energy.
Q̇ − Ẇ = ṁ [(h₂ − h₁) + (V₂² − V₁²) / 2000 + g(z₂ − z₁) / 1000] Δke = (V₂² − V₁²) / 2000 kJ/kg Δpe = g(z₂ − z₁) / 1000 kJ/kg Δe = (h₂ − h₁) + Δke + ΔpeVariables: Q̇ is heat transfer rate, Ẇ is shaft work rate, ṁ is mass flow rate, h is specific enthalpy, V is velocity, z is elevation, and g is gravitational acceleration.
Unit note: Enthalpy is entered in kJ/kg. Kinetic and potential terms are converted from J/kg to kJ/kg, so the final rate form reports kW.
How to use this calculator
- Select the target you want to solve for, such as shaft work rate or outlet enthalpy.
- Enter the known steady flow data for the device, including mass flow, enthalpy levels, velocities, and elevations.
- Use positive heat values for energy entering the control volume and positive shaft work for output from the device.
- Press the calculate button to display the result above the form, directly below the header area.
- Review the summary metrics, balance residual, and device interpretation to confirm the physical meaning of the solution.
- Download the calculated results or example table as CSV or PDF for reporting, lab work, or design notes.
Frequently asked questions
1. What does the steady flow energy equation describe?
It relates heat transfer and shaft work to changes in enthalpy, kinetic energy, and potential energy for a control volume operating at steady conditions.
2. When is the steady flow assumption appropriate?
Use it when mass and energy inside the control volume do not change with time, even though fluid continuously enters and leaves.
3. Why is enthalpy included instead of internal energy alone?
Enthalpy combines internal energy with flow work, making it the natural property for open systems where fluid crosses the control surface.
4. How should I assign signs to heat and work?
Enter heat as positive when it enters the device. Enter shaft work as positive when the device delivers work to the surroundings.
5. Can this calculator model turbines and compressors?
Yes. Turbines usually return positive shaft work, while compressors often require negative shaft work because power is supplied to the device.
6. Why do velocity and elevation matter?
They capture kinetic and potential energy changes. These terms are often small, but they can be important in nozzles, diffusers, and tall installations.
7. What does a residual near zero mean?
It means the entered data and the computed result satisfy the energy balance within rounding, which improves confidence in the calculation.
8. Can I use different unit systems here?
This page is set for SI-style inputs. Convert other units first so enthalpy, velocity, elevation, and rates remain consistent.